(1 + y) dx - (1 - x) dy = 0(1 + y) dx = (1 - x) dy[1/(1 - x)]dx =[1 /(1 + y)]dyd(ln(1 - x))=d(ln(1 + y))ln(1 - x) + C1 = ln(1 + y) （C1為任意實(shí)數(shù)）1 + y = e^[ln(1 - x) + C1]y = C*(1 - x) - 1 (C為任意...答案后面是加1不是減1。。。。。題目也不對(duì)啊(1-x)dx-(1+y)dy=0求通解不是(1 + y) dx - (1 - x) dy防造這樣做為啥百度的答案是一個(gè)類似于圓的方程不知道。。。。你的答案確定嗎嘿嘿 不確定(1 + y) dx - (1 - x) dy = 0？(1 + y) dx = (1 - x) dy[1/(1 - x)]dx =[1 /(1 + y)]dyd(ln(1 - x))=d(ln(1 + y))ln(1 - x) + C1 = ln(1 + y) （C1為任意實(shí)數(shù)）1 + y = e^[ln(1 - x) + C1]y = C*(1 - x) - 1 (C為任意實(shí)數(shù)）我去打問號(hào)的是啥意思你這是初一的題吧大一。。。。。還初一-_-||不會(huì)微分方程的通解-_-||抱歉沒事盡力而為